Edexcel Paper 2 2021 October — Question 9 3 marks

Exam BoardEdexcel
ModulePaper 2 (Paper 2)
Year2021
SessionOctober
Marks3
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicSequences and series, recurrence and convergence
TypeGeometric series with summation
DifficultyStandard +0.8 This requires recognizing that cos(180n)° alternates between +1 and -1, converting the sum to a geometric series with ratio -3/4, applying the geometric series formula with correct starting index n=2, and careful algebraic manipulation. It combines trigonometric insight with series techniques, going beyond routine application.
Spec1.04j Sum to infinity: convergent geometric series |r|<11.05a Sine, cosine, tangent: definitions for all arguments
  1. Show that
$$\sum _ { n = 2 } ^ { \infty } \left( \frac { 3 } { 4 } \right) ^ { n } \cos ( 180 n ) ^ { \circ } = \frac { 9 } { 28 }$$
VI4V SIHI NI SIIIM ION OCVIAV SIHI NI III IM I ON OCVJ4V SIHI NI IMIMM ION OC
Question 9:
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(a = \left(\frac{3}{4}\right)^2\) or \(a = \frac{9}{16}\), or \(r = -\frac{3}{4}\)B1 2.2a — Deduces correct first term or common ratio
\(\sum_{n=2}^{\infty}\left(\frac{3}{4}\right)^n\cos(180n)^\circ = \dfrac{\frac{9}{16}}{1-\left(-\frac{3}{4}\right)} = \ldots\)M1 3.1a — Recognises infinite geometric series; applies sum to infinity formula with \(a=\frac{9}{16}\), \(r=\pm\frac{3}{4}\)
\(= \frac{9}{28}\)A1* 1.1b — Correct proof
Alternative 1:
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(\sum_{n=1}^{\infty}\left(\frac{3}{4}\right)^n\cos(180n)^\circ = \dfrac{-\frac{3}{4}}{1-\left(-\frac{3}{4}\right)} = \ldots\) or \(r=-\frac{3}{4}\)B1 2.2a — Deduces correct sum to infinity (from \(n=1\)) or common ratio
\(\sum_{n=2}^{\infty}\left(\frac{3}{4}\right)^n\cos(180n)^\circ = -\frac{3}{7} - \left(-\frac{3}{4}\right)\)M1 3.1a — Calculates required value by subtracting first term from sum to infinity
\(= \frac{9}{28}\)A1* 1.1b — Correct proof
Alternative 2:
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(\sum_{n=2}^{\infty}\left(\frac{3}{4}\right)^n\cos(180n)^\circ = \left(\frac{3}{4}\right)^2 - \left(\frac{3}{4}\right)^3 + \left(\frac{3}{4}\right)^4 - \ldots\)B1 2.2a — Deduces correct first term or common ratio
\(\left(\frac{3}{4}\right)^2+\left(\frac{3}{4}\right)^4+\ldots = \left(\frac{3}{4}\right)^2\left(\dfrac{1}{1-\left(\frac{3}{4}\right)^2}\right)\) or \(-\left(\frac{3}{4}\right)^3-\left(\frac{3}{4}\right)^5-\ldots = -\left(\frac{3}{4}\right)^3\left(\dfrac{1}{1-\left(\frac{3}{4}\right)^2}\right)\)M1 3.1a — Splits into "odds" and "evens", attempts sum of both parts
\(= \frac{9}{28}\)A1* 1.1b — Correct proof
Alternative 3:
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(S = \left(\frac{3}{4}\right)^2 - \left(\frac{3}{4}\right)^3 + \left(\frac{3}{4}\right)^4 - \ldots\)B1 2.2a — Deduces correct first term
\(= \left(\frac{3}{4}\right)^2\left(1 - \frac{3}{4} + \left(\frac{3}{4}\right)^2 - \ldots\right) = \left(\frac{3}{4}\right)^2\left(\frac{1}{4}+S\right) \Rightarrow \frac{7}{16}S = \frac{9}{64}\)M1 3.1a — Takes out first term, expresses RHS in terms of \(S\), rearranges
\(= \frac{9}{28}\)A1* 1.1b — Correct proof 
# Question 9:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $a = \left(\frac{3}{4}\right)^2$ or $a = \frac{9}{16}$, or $r = -\frac{3}{4}$ | B1 | 2.2a — Deduces correct first term or common ratio |
| $\sum_{n=2}^{\infty}\left(\frac{3}{4}\right)^n\cos(180n)^\circ = \dfrac{\frac{9}{16}}{1-\left(-\frac{3}{4}\right)} = \ldots$ | M1 | 3.1a — Recognises infinite geometric series; applies sum to infinity formula with $a=\frac{9}{16}$, $r=\pm\frac{3}{4}$ |
| $= \frac{9}{28}$ | A1* | 1.1b — Correct proof |

**Alternative 1:**

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\sum_{n=1}^{\infty}\left(\frac{3}{4}\right)^n\cos(180n)^\circ = \dfrac{-\frac{3}{4}}{1-\left(-\frac{3}{4}\right)} = \ldots$ or $r=-\frac{3}{4}$ | B1 | 2.2a — Deduces correct sum to infinity (from $n=1$) or common ratio |
| $\sum_{n=2}^{\infty}\left(\frac{3}{4}\right)^n\cos(180n)^\circ = -\frac{3}{7} - \left(-\frac{3}{4}\right)$ | M1 | 3.1a — Calculates required value by subtracting first term from sum to infinity |
| $= \frac{9}{28}$ | A1* | 1.1b — Correct proof |

**Alternative 2:**

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\sum_{n=2}^{\infty}\left(\frac{3}{4}\right)^n\cos(180n)^\circ = \left(\frac{3}{4}\right)^2 - \left(\frac{3}{4}\right)^3 + \left(\frac{3}{4}\right)^4 - \ldots$ | B1 | 2.2a — Deduces correct first term or common ratio |
| $\left(\frac{3}{4}\right)^2+\left(\frac{3}{4}\right)^4+\ldots = \left(\frac{3}{4}\right)^2\left(\dfrac{1}{1-\left(\frac{3}{4}\right)^2}\right)$ or $-\left(\frac{3}{4}\right)^3-\left(\frac{3}{4}\right)^5-\ldots = -\left(\frac{3}{4}\right)^3\left(\dfrac{1}{1-\left(\frac{3}{4}\right)^2}\right)$ | M1 | 3.1a — Splits into "odds" and "evens", attempts sum of both parts |
| $= \frac{9}{28}$ | A1* | 1.1b — Correct proof |

**Alternative 3:**

| Answer/Working | Mark | Guidance |
|---|---|---|
| $S = \left(\frac{3}{4}\right)^2 - \left(\frac{3}{4}\right)^3 + \left(\frac{3}{4}\right)^4 - \ldots$ | B1 | 2.2a — Deduces correct first term |
| $= \left(\frac{3}{4}\right)^2\left(1 - \frac{3}{4} + \left(\frac{3}{4}\right)^2 - \ldots\right) = \left(\frac{3}{4}\right)^2\left(\frac{1}{4}+S\right) \Rightarrow \frac{7}{16}S = \frac{9}{64}$ | M1 | 3.1a — Takes out first term, expresses RHS in terms of $S$, rearranges |
| $= \frac{9}{28}$ | A1* | 1.1b — Correct proof |

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\begin{enumerate}
  \item Show that
\end{enumerate}

$$\sum _ { n = 2 } ^ { \infty } \left( \frac { 3 } { 4 } \right) ^ { n } \cos ( 180 n ) ^ { \circ } = \frac { 9 } { 28 }$$

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VI4V SIHI NI SIIIM ION OC & VIAV SIHI NI III IM I ON OC & VJ4V SIHI NI IMIMM ION OC \\
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\hfill \mbox{\textit{Edexcel Paper 2 2021 Q9 [3]}}
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